Download Note An Upper Bound for the Transversal Numbers

JOURNAL
OF COMBINATORIAL
Series B SO, 129-133
THEORY,
(1990)
Note
An Upper Bound for the Transversal
Numbers of 4-Uniform
Hypergraphs’
FENG-CHU
Department
LAI
of Applied Mathematics.
National
Hsinehu 30050, Taiwan, Republic
Chiao Tung University,
of China
GERARD J. CHANG
Department
of Applied Mathematics.
National
Chiao Tung University,
Hsinchu 30050, Taiwan, Republic of China, and
Institute of Informaiion
Science, Academia Sinica,
Nankang,
Taipei 11529, Taiwan, Republic of China
Communicated
Received
by the Editors
April
11, 1988
The main purpose of this paper is to prove that if H is a 4-uniform
with n vertices
and m edges, then the transversal
number
r(H)
0 1990 Academic
All
hypergraph
<2(m +n)/9.
Press, Inc.
standard
terminology
of hypergraphs is from [ 11. Suppose
hypergraph with n vertices and m edges. Tuza
[2] proposed the problem of finding an upper bound for the transversal
number r(H), of the form t(H) < ck(n + m), where ck depends only on k.
More precisely, we want to determine ck z sup z(H)/(m + n), where H runs
over all k-uniform hypergraphs of n vertices and m edges. It is easy to see
that c1 = 4 and c2 = i. Tuza [2] proved that c3 = $ and asked if ck is
0(1/k).
For any positive integer p we can construct a k-uniform hypergraph H
of n = k + p vertices xi, .... x, and m = [n/p1 edges e,, .... e,, where
H = (V, E) is a k-uniform
ej= {x,
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130
LA1 AND
CHANG
and
e,
= {Xl)
. ..) Xk}.
Then r(H) = 2. To make r(H)/(n + m) large, we only have to find p
such that n + m = k + 1 + p + [k/p] is minimum.
It is an easy exercise
to check that n + m achieves minimum
at p = LJif].
So ck B bk =
2/(k + 1 + L$j
+ rk/LkJ).
Note that ck = b, = l/(k + 1) for k < 3. The
main purpose of this paper is to prove that cq = b, = $. In fact we prove a
more general theorem.
THEOREM.
Suppose H = (V, E) is a 4-uniform hypergraph of n vertices
and m edges. Zf H has t end edges, which are edges containing vertices of
degree one, then z(H) < (2n + 2m - t)/9.
ProoJ
Without loss of generality we may assume that H is connected.
We shall prove the theorem by induction on m. The theorem is trivial for
m=O since z(H)=t=O.
Suppose X= {vi, .... tak} c V is a set of k vertices. Denote by H’ the
hypergraph obtained from H by deleting all edges e, , .... eP that meet X and
ail vertices only in e, u . . . u eP. Then
z(H) d k + z( H’) < (2n’ + 2m’ - t’ + 9k)/9,
n’=n-r,
m’=m-p,
t’=t-q+s,
where r is the number of vertices only in e, u . . . u eP, q is the number of
end edges in e,, .... eP, and s is the number of end edges in H’ which are
not end edges in H. Consequently,
T(H) < (2n + 2m - t)/9 - (2p + 2r + s - q - 9k)/9.
So the theorem holds when A* - 2p + 2r + s - q 2 91x1. In the case when
each vertex in X is of degree 2 2, r > q + k and the theorem holds when
A=2p+r+s>81XI.
Case 1. If there is a vertex x of degree p 2 4, then choose X= {x}.
pa4 implies A>81XI.
Case 2. Suppose all vertices of H are of degree 3. Choose a k-cycle
C= (x1, el, x2, f3, .... xk, ek, x, ) of minimum length. Denote by e; the edge
other than e,-i and ei that contains xi.
TRANSVERSAL
NUMBERS
131
OF HYPERGRAPHS
e2
X
f?
FIGURE
e’
FIGURE
1
2
2
e2
132
LA1
AND
CHANG
In the case k=2, let X=(x,}.
Thenp=3;
r>2 when e;=e;;
ral
and
s>l when e;#e;. Thus d>8lXI.
Now assumek 2 3. Choose a vertex x E ek~ f - { xk _ i, x,}; denote by e’
and e” the edges other than eke, that contain x. By the minimality of the
length of C, the edges e,, .... ek, e;, .... eh, e’, err are distinct. Now consider
x= {Xl, x3, x5, .... X& 1} when k is even and X = {x,, x3, .... xkm2, x}
when k is odd. Then p = 3k/2 and r, s > /XI = k/2 when k is even (see Fig. 1
for k = 6); and p = 3(k + 1)/2 and r, s 3 1x1= (k + 1)/2 when k is odd (see
Fig. 2 for k = 5). In any case d >, 81X(.
Case 3. Suppose all vertices are of degree at most 3 and there is a
vertex x of degree 3 contained in e,, e,, e3; but e, has at least one vertex
of degree 6 2.
If e, is an end edge, then consider X= {x}. p = 3 and r > 2 imply
A > 81x1. So we can assumethat e, is not an end edge. Suppose y is a vertex of degree 2 in e,, say yee, ne. If e=e2 or e3, then consider X= {x}.
Againp=3
and r>2 imply A>81XI. So assumee#e, and e#e,. If e is
not an end edge, then consider X= ix}. p = 3, r 2 1 and s B 1 imply
A 2 81x1.
Now we can assume e = (y, y,, 112,y3} with 1 = deg(y,) ddeg(y,) 6
deg(y,)<3.
If deg(y,)=3,
then choose X= {y3} to get A>8lXI.
So
assumedeg( u3) = 2, say y, E e n e’. In the caseof deg( yz) = 1 or e’ is an end
edge we consider X= { y3}, then p = 2, r > 3, and s > 1; in the case of
deg(y,) = 2 and e’ is not an end edge we consider X= { y2}, then p = 2,
r>2, and ~22. In any case A~8lXl.
Case 4. Suppose all vertices of H are of degree at most 2. Suppose
there is an end edge e containing at least two vertices of degree one.
Choose X= {x}, where x is a vertex in e of maximum degree p. Then
either p=q=l
and r=4, which imply d*>9=9lXI;
or else p=2
and r>q+233
which imply A*=2p+2r-q+s>2p+2q+4-q+sa
9 = 91x1. So we can assumethat every and edge has exactly one vertex of
degree one. Count the degrees of all vertices. There are t vertices of degree
one and n - t of degree 2; then we have 4m = t + 2(n - t) = 2n - t.
Suppose there are two distinct edges e and f such that le n f 1= 3, say
e={x,,x,,x,,x,}
andf=( x1, x2, x3, x5}. If deg(x,) = 1 or deg(x,) = 1,
then we choose X= (xi}. In this case p = 2 and r 2 4, which imply
Aa8lX(.
So we may assume deg(x,)=deg(x,)=2,
say x4eene’
and
x5 Ef n f ‘. If e’ or f' is not an edge, then we choose X= (xi }. In this case
p = 2, r = 3, and s 2 1, which imply A 3 8 [XI. So we may assumethat e’ and
f’ are end edges, say e’= (x4,~,,y2,y3}
with deg(y,)= 1 and deg(y,)=
deg(y,) = 2 (since every end edge has exactly one vertex of degree one). Let
y1 Ee’ n e, and y, Ee’ n e2. If e, or e2 is an end edge, we can without loss
of generality assumethat e, is an end edge. Now choose X= { JJ,}. Then
TRANSVERSAL NUMBERS OF HYPERGRAPHS
133
either p = 2, r = 3, and s > 1 (when e, is an end edge), or else p = 2, r = 2,
and s3 2 (when e, is not an end edge). In any case A > 8. So we can
assume that Jen fl d 2 for any two distinct edges e and J:
Choose maximum number of vertices xr , .... xk of degree 2, say xi E ej n f,
for 1 d id k, such that e,, fr, e,, fi, .... ek, fk are distinct. Let g1 , .... g,- Zk
be the remaining edges of E. By the maximality
of k, the edges gj are
pairwise disjoint. Each edge gj has at least three vertices of degree 2, which
are also in some e,. or f,; call such vertices common vertices. Note that there
are at least 3(m - 2k) common vertices. On the other hand, e, u fj has at
most three common vertices for any i. Otherwise the fact that any two
distinct edges intersect at no more than two vertices implies that either
lej n f, 1= lei n gj 1= (f, n gr 1= 2 for some gj or else there are common vertices x7 E e, A gj and x,* * Ef, n gj, for distinct j and j’. For the former case,
H has exactly 3 edges and 6 vertices and t(H) = 2, so the theorem holds.
For the latter case, we can replace xi by XT and x,+* to get k + 1 vertices
whose containing edges are distinct, in contradiction to the maximality of
k. Hence 3(m - 2k) < 3k, i.e., m d 3k. Since x1, .... xk together with a vertex
in gj for 1 < j < m - 2k form a transversal,
r(H)<k+m-2k
d 2mJ3
(since m < 3k)
= (2m + 2n - t)/9
(since 4m = 2n - t).
REFERENCES
1. C. BERGE,“Graphs and Hypergraphs”, North-Holland,
Z. TUZA, Covering all cliques of a graph, preprint.
2.
Amsterdam, 1973.
Q.E.D.